Study of the function x3 3 x2 8x 3. Problems from the collection of Kuznetsov L

When plotting function graphs, it is useful to adhere to the following plan:

1. Find the domain of definition of the function and determine the discontinuity points, if any.

2. Determine whether the function is even or odd or neither. If the function is even or odd, then it is enough to consider its values ​​at x>0, and then symmetrically with respect to the OY axis or the origin of coordinates, restore it for the values x<0 .

3. Examine the function for periodicity. If the function is periodic, then it is enough to consider it on one period.

4. Find the intersection points of the function graph with the coordinate axes (if possible)

5. Conduct a study of the function at the extremum and find the intervals of increase and decrease of the function.

6. Find the inflection points of the curve and the intervals of convexity and concavity of the function.

7. Find the asymptotes of the graph of the function.

8. Using the results of steps 1-7, construct a graph of the function. Sometimes several additional points are found for greater accuracy; their coordinates are calculated using the equation of the curve.

Example. Explore function y=x 3 -3x and build a graph.

1) The function is defined on the interval (-∞; +∞). There are no breaking points.

2) The function is odd, because f(-x) = -x 3 -3(-x) = -x 3 +3x = -f(x), therefore, it is symmetrical about the origin.

3) The function is not periodic.

4) Points of intersection of the graph with the coordinate axes: x 3 -3x=0, x = , x = -, x = 0, those. the graph of the function intersects the coordinate axes at the points: ( ; 0 ), (0; 0 ), (-; 0 ).

5) Find possible extremum points: y′ = 3x 2 -3; 3x 2 -3=0; x =-1; x = 1. The domain of definition of the function will be divided into intervals: (-∞; -1), (-1; 1), (1; +∞). Let's find the signs of the derivative in each resulting interval:

On the interval (-∞; -1) y′>0 – function increases

On the interval (-1; 1) y′<0 – the function is decreasing

On the interval (1; +∞) y′>0 – function increases. Dot x =-1 – maximum point; x = 1 – minimum point.

6) Find the inflection points: y′′ = 6x; 6x = 0; x = 0. Dot x = 0 divides the domain of definition into intervals (-∞; 0), (0; +∞). Let's find the signs of the second derivative in each resulting interval:

On the interval (-∞;0) y′′<0 – the function is convex

On the interval (0; +∞) y′′>0 – function is concave. x = 0– inflection point.

7) The graph has no asymptotes

8) Let's build a graph of the function:

Example. Explore the function and construct its graph.

1) The domain of definition of the function is the intervals (-¥; -1) È (-1; 1) È (1; ¥). Range of values of this function is the interval (-¥; ¥).

The break points of the function are the points x = 1, x = -1.

2) The function is odd, because .

3) The function is not periodic.

4) The graph intersects the coordinate axes at the point (0; 0).

5) Find critical points.

Critical points: x = 0; x = -; x = ; x = -1; x = 1.

Find the intervals of increasing and decreasing functions. To do this, we determine the signs of the derivative of the function on intervals.

-¥ < x< -, y¢> 0, the function is increasing

-< x < -1, y¢ < 0, функция убывает

1 < x < 0, y¢ < 0, функция убывает

0 < x < 1, y¢ < 0, функция убывает

1 < x < , y¢ < 0, функция убывает

< x < ¥, y¢ > 0, the function increases

It is clear that the point X= -is the maximum point, and the point X= is the minimum point. The function values ​​at these points are equal to 3/2 and -3/2, respectively.

6) Find the second derivative of the function

Oblique asymptote equation: y = x.

8) Let's build a graph of the function.

If the problem requires a complete study of the function f (x) = x 2 4 x 2 - 1 with the construction of its graph, then we will consider this principle in detail.

To solve the problem of this type properties and graphs of the main elementary functions. The research algorithm includes the following steps:

Finding the domain of definition

Since research is carried out on the domain of definition of the function, it is necessary to start with this step.

Example 1

The given example involves finding the zeros of the denominator in order to exclude them from the ODZ.

4 x 2 - 1 = 0 x = ± 1 2 ⇒ x ∈ - ∞ ; - 1 2 ∪ - 1 2 ; 1 2 ∪ 1 2 ; +∞

As a result, you can get roots, logarithms, and so on. Then the ODZ can be searched for a root of an even degree of type g (x) 4 by the inequality g (x) ≥ 0, for the logarithm log a g (x) by the inequality g (x) > 0.

Studying the boundaries of the ODZ and finding vertical asymptotes

There are vertical asymptotes at the boundaries of the function, when the one-sided limits at such points are infinite.

Example 2

For example, consider the border points equal to x = ± 1 2.

Then it is necessary to study the function to find the one-sided limit. Then we get that: lim x → - 1 2 - 0 f (x) = lim x → - 1 2 - 0 x 2 4 x 2 - 1 = = lim x → - 1 2 - 0 x 2 (2 x - 1 ) (2 x + 1) = 1 4 (- 2) · - 0 = + ∞ lim x → - 1 2 + 0 f (x) = lim x → - 1 2 + 0 x 2 4 x - 1 = = lim x → - 1 2 + 0 x 2 (2 x - 1) (2 x + 1) = 1 4 (- 2) (+ 0) = - ∞ lim x → 1 2 - 0 f (x) = lim x → 1 2 - 0 x 2 4 x 2 - 1 = = lim x → 1 2 - 0 x 2 (2 x - 1) (2 x + 1) = 1 4 (- 0) 2 = - ∞ lim x → 1 2 - 0 f (x) = lim x → 1 2 - 0 x 2 4 x 2 - 1 = = lim x → 1 2 - 0 x 2 (2 x - 1) (2 x + 1) = 1 4 ( + 0) 2 = + ∞

This shows that the one-sided limits are infinite, which means the straight lines x = ± 1 2 are the vertical asymptotes of the graph.

Study of a function and whether it is even or odd

When the condition y (- x) = y (x) is satisfied, the function is considered even. This suggests that the graph is located symmetrically with respect to Oy. When the condition y (- x) = - y (x) is satisfied, the function is considered odd. This means that the symmetry is relative to the origin of coordinates. If at least one inequality is not satisfied, we obtain a function of general form.

The equality y (- x) = y (x) indicates that the function is even. When constructing, it is necessary to take into account that there will be symmetry with respect to Oy.

To solve the inequality, intervals of increasing and decreasing are used with the conditions f " (x) ≥ 0 and f " (x) ≤ 0, respectively.

Definition 1

Stationary points- these are the points that turn the derivative to zero.

Critical points- these are internal points from the domain of definition where the derivative of the function is equal to zero or does not exist.

When making a decision, the following notes must be taken into account:

• for existing intervals of increasing and decreasing inequalities of the form f " (x) > 0, critical points are not included in the solution;
• points at which the function is defined without a finite derivative must be included in the intervals of increasing and decreasing (for example, y = x 3, where the point x = 0 makes the function defined, the derivative has the value of infinity at this point, y " = 1 3 x 2 3, y "(0) = 1 0 = ∞, x = 0 is included in the increasing interval);
• To avoid disagreements, it is recommended to use mathematical literature recommended by the Ministry of Education.

Inclusion of critical points in intervals of increasing and decreasing if they satisfy the domain of definition of the function.

Definition 2

For determining the intervals of increase and decrease of a function, it is necessary to find:

• derivative;
• critical points;
• divide the definition domain into intervals using critical points;
• determine the sign of the derivative on each of the intervals, where + is an increase and - is a decrease.

Example 3

Find the derivative on the domain of definition f " (x) = x 2 " (4 x 2 - 1) - x 2 4 x 2 - 1 " (4 x 2 - 1) 2 = - 2 x (4 x 2 - 1) 2 .

Solution

To solve you need:

• find stationary points, this example has x = 0;
• find the zeros of the denominator, the example takes the value zero at x = ± 1 2.

We place points on the number line to determine the derivative on each interval. To do this, it is enough to take any point from the interval and perform a calculation. If the result is positive, we depict + on the graph, which means the function is increasing, and - means it is decreasing.

For example, f " (- 1) = - 2 · (- 1) 4 - 1 2 - 1 2 = 2 9 > 0, which means that the first interval on the left has a + sign. Consider on the number line.

Answer:

• the function increases on the interval - ∞; - 1 2 and (- 1 2 ; 0 ] ;
• there is a decrease in the interval [ 0 ; 1 2) and 1 2 ; + ∞ .

In the diagram, using + and -, the positivity and negativity of the function are depicted, and the arrows indicate decrease and increase.

Extremum points of a function are points where the function is defined and through which the derivative changes sign.

Example 4

If we consider an example where x = 0, then the value of the function in it is equal to f (0) = 0 2 4 · 0 2 - 1 = 0. When the sign of the derivative changes from + to - and passes through the point x = 0, then the point with coordinates (0; 0) is considered the maximum point. When the sign changes from - to +, we obtain a minimum point.

Convexity and concavity are determined by solving inequalities of the form f "" (x) ≥ 0 and f "" (x) ≤ 0. Less commonly used is the name convexity down instead of concavity, and convexity upward instead of convexity.

Definition 3

For determining the intervals of concavity and convexity necessary:

• find the second derivative;
• find the zeros of the second derivative function;
• divide the definition area into intervals with the appearing points;
• determine the sign of the interval.

Example 5

Find the second derivative from the domain of definition.

Solution

f "" (x) = - 2 x (4 x 2 - 1) 2 " = = (- 2 x) " (4 x 2 - 1) 2 - - 2 x 4 x 2 - 1 2 " (4 x 2 - 1) 4 = 24 x 2 + 2 (4 x 2 - 1) 3

We find the zeros of the numerator and denominator, where in our example we have that the zeros of the denominator x = ± 1 2

Now you need to plot the points on the number line and determine the sign of the second derivative from each interval. We get that

Answer:

• the function is convex from the interval - 1 2 ; 12 ;
• the function is concave from the intervals - ∞ ; - 1 2 and 1 2; + ∞ .

Definition 4

Inflection point– this is a point of the form x 0 ; f (x 0) . When it has a tangent to the graph of the function, then when it passes through x 0 the function changes sign to the opposite.

In other words, this is a point through which the second derivative passes and changes sign, and at the points themselves it is equal to zero or does not exist. All points are considered to be the domain of the function.

In the example, it was clear that there are no inflection points, since the second derivative changes sign while passing through the points x = ± 1 2. They, in turn, are not included in the scope of definition.

Finding horizontal and oblique asymptotes

When defining a function at infinity, you need to look for horizontal and oblique asymptotes.

Definition 5

Oblique asymptotes are depicted using straight lines, given by the equation y = k x + b, where k = lim x → ∞ f (x) x and b = lim x → ∞ f (x) - k x.

For k = 0 and b not equal to infinity, we find that the oblique asymptote becomes horizontal.

In other words, asymptotes are considered to be lines to which the graph of a function approaches at infinity. This facilitates quick construction of a function graph.

If there are no asymptotes, but the function is defined at both infinities, it is necessary to calculate the limit of the function at these infinities in order to understand how the graph of the function will behave.

Example 6

Let's consider as an example that

k = lim x → ∞ f (x) x = lim x → ∞ x 2 4 x 2 - 1 x = 0 b = lim x → ∞ (f (x) - k x) = lim x → ∞ x 2 4 x 2 - 1 = 1 4 ⇒ y = 1 4

is a horizontal asymptote. After examining the function, you can begin to construct it.

Calculating the value of a function at intermediate points

To make the graph more accurate, it is recommended to find several function values ​​at intermediate points.

Example 7

From the example we considered, it is necessary to find the values ​​of the function at the points x = - 2, x = - 1, x = - 3 4, x = - 1 4. Since the function is even, we get that the values ​​coincide with the values ​​at these points, that is, we get x = 2, x = 1, x = 3 4, x = 1 4.

Let's write and solve:

F (- 2) = f (2) = 2 2 4 2 2 - 1 = 4 15 ≈ 0, 27 f (- 1) - f (1) = 1 2 4 1 2 - 1 = 1 3 ≈ 0 , 33 f - 3 4 = f 3 4 = 3 4 2 4 3 4 2 - 1 = 9 20 = 0 , 45 f - 1 4 = f 1 4 = 1 4 2 4 1 4 2 - 1 = - 1 12 ≈ - 0.08

To determine the maxima and minima of the function, inflection points, and intermediate points, it is necessary to construct asymptotes. For convenient designation, intervals of increasing, decreasing, convexity, and concavity are recorded. Let's look at the picture below.

It is necessary to draw graph lines through the marked points, which will allow you to approach the asymptotes by following the arrows.

This concludes the full exploration of the function. There are cases of constructing some elementary functions for which geometric transformations are used.

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Solver Kuznetsov.
III Charts

Task 7. Conduct a complete study of the function and construct its graph.

        Before you start downloading your options, try solving the problem according to the example given below for option 3. Some of the options are archived in .rar format

        7.3 Conduct a full study of the function and plot it

Solution.

        1) Scope of definition:         or        , that is        .
.
Thus:         .

        2) There are no points of intersection with the Ox axis. Indeed, the equation         has no solutions.
There are no points of intersection with the Oy axis, since        .

        3) The function is neither even nor odd. There is no symmetry about the ordinate axis. There is also no symmetry about the origin. Because
.
We see that         and        .

        4) The function is continuous in the domain of definition
.

; .

; .
Consequently, the point         is a point of discontinuity of the second kind (infinite discontinuity).

5) Vertical asymptotes:       

Let's find the oblique asymptote        . Here

;
.
Therefore, we have horizontal asymptote: y=0. Oblique asymptotes No.

        6) Let’s find the first derivative. First derivative:
.
And that's why
.
Let's find stationary points where the derivative is equal to zero, that is
.

        7) Let’s find the second derivative. Second derivative:
.
And this is easy to verify, since

This lesson covers the topic "Investigation of a function and related problems." This lesson covers graphing functions using derivatives. The function is studied, its graph is constructed and a number of related problems are solved.

Topic: Derivative

Lesson: Exploring a Functionand related tasks

It is necessary to study this function, construct a graph, find intervals of monotonicity, maximums, minimums, and what problems accompany knowledge about this function.

First, let's take full advantage of the information provided by the function without derivative.

1. Find the intervals of constant sign of the function and construct a sketch of the graph of the function:

1) Let's find.

2) Function roots: , from here

3) Intervals of constant sign of the function (see Fig. 1):

Rice. 1. Intervals of constant sign of a function.

Now we know that in the interval and the graph is above the X-axis, in the interval - below the X-axis.

2. Let's build a graph in the vicinity of each root (see Fig. 2).

Rice. 2. Graph of a function in the vicinity of the root.

3. Construct a graph of the function in the vicinity of each discontinuity point in the domain of definition. The domain of definition breaks at the point . If the value is close to the point, then the value of the function tends to (see Fig. 3).

Rice. 3. Graph of the function in the vicinity of the discontinuity point.

4. Let us determine how the graph behaves in the vicinity of points at infinity:

Let's write it using limits

. It is important that for very large values, the function is almost no different from unity.

Let's find the derivative, intervals of its constant sign and they will be intervals of monotonicity for the function, find those points at which the derivative is equal to zero, and find out where the maximum point is and where the minimum point is.

From here, . These points are internal points of the domain of definition. Let's find out what sign of the derivative is on the intervals, and which of these points is the maximum point and which is the minimum point (see Fig. 4).

Rice. 4. Intervals of constant sign of the derivative.

From Fig. 4 it can be seen that the point is a minimum point, the point is a maximum point. The value of the function at the point is . The value of the function at the point is 4. Now let's build a graph of the function (see Fig. 5).

Rice. 5. Function graph.

Thus we built graph of a function. Let's describe it. Let us write down the intervals over which the function decreases monotonically: , are those intervals where the derivative is negative. The function increases monotonically on the intervals and . - minimum point, - maximum point.

Find the number of roots of the equation depending on the parameter values.

1. Construct a graph of the function. The graph of this function is plotted above (see Fig. 5).

2. Dissect the graph with a family of straight lines and write down the answer (see Fig. 6).

Rice. 6. Intersection of the graph of a function with straight lines.

1) When - one solution.

2) When - two solutions.

3) When - three solutions.

4) When - two solutions.

5) When - three solutions.

6) When - two solutions.

7) When - one solution.

Thus, we solved one of the important problems, namely, finding the number of solutions to the equation depending on the parameter . There may be different special cases, for example, in which there will be one solution, or two solutions, or three solutions. Note that these special cases, all answers to these special cases are contained in the general answer.

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Additional web resources

2. Portal Natural Sciences ().

Make it at home

No. 45.7, 45.10 (Algebra and the beginnings of analysis, grade 10 (in two parts). Problem book for general education institutions (profile level) edited by A. G. Mordkovich. - M.: Mnemosyne, 2007.)